Declaration and Access Control The object is created with new keyword At Compile-time At run-time None of these Depends on the code At Compile-time At run-time None of these Depends on the code ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What is the result of compiling and running the following code?public class Tester{static int x = 4;int y = 9; public Tester(){System.out.print(this.x); // line 1printVariables();}public static void printVariables(){System.out.print(x); // line 2System.out.print(y); // line 3}public static void main(String... args) { // line 4new Tester();}} Compile error at line 2 (must access x by writing Tester.x) Compile error at line 3 (static methods can't make reference to non-static variables) Compile error at line 1 (static x must be only accessed inside static methods) Compile error at line 4 (invalid argument type for method main) 49 Compile error at line 2 (must access x by writing Tester.x) Compile error at line 3 (static methods can't make reference to non-static variables) Compile error at line 1 (static x must be only accessed inside static methods) Compile error at line 4 (invalid argument type for method main) 49 ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control Choose all the lines which if inserted independently instead of "//insert code here" will allow the following code to compile:public class Test{ public static void main(String args[]){ add(); add(1);add(1, 2); }// insert code here} static void add(int...args){} void add(Integer... args){} static void add(int args...){} static void add(int[]... args){} static void add(int... args, int y){} static void add(int...args){} void add(Integer... args){} static void add(int args...){} static void add(int[]... args){} static void add(int... args, int y){} ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What will be the output?public class Test{ public static void main(String[] args){ String value = "abc"; changeValue(value); System.out.println(value); } public static void changeValue(String a){ a = "xyz"; }} abc Compilation clean but no output xyz Compilation fails None of these abc Compilation clean but no output xyz Compilation fails None of these ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control Suppose a class has public visibility. In this class we define a protected method. Which of the following statements is correct? In a class, you cannot declare methods with a lower visibility than the visibility of the class in which it is defined. From within protected methods you do not have access to public methods. This method is only accessible from inside the class itself and from inside all subclasses. This method is accessible from within the class itself and from within all classes defined in the same package as the class itself. In a class, you cannot declare methods with a lower visibility than the visibility of the class in which it is defined. From within protected methods you do not have access to public methods. This method is only accessible from inside the class itself and from inside all subclasses. This method is accessible from within the class itself and from within all classes defined in the same package as the class itself. ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What will be the output for the below code?static public class Test{ public static void main(String[] args){ char c = 'a'; switch(c){ case 65 : System.out.println("one");break; case 'a': System.out.println("two");break; case 3 : System.out.println("three"); } }} Compile error - char can't be permitted in switch statement. one None of these two Compile error - Illegal modifier for the class Test; only public, abstract & final are permitted. Compile error - char can't be permitted in switch statement. one None of these two Compile error - Illegal modifier for the class Test; only public, abstract & final are permitted. ANSWER DOWNLOAD EXAMIANS APP
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